Question: $ \int_{-1}^1 \int_{1 - \sqrt{1 - y^2}}^{1 + \sqrt{1 - y^2}} dx \, dy$ Switch the bounds of the double integral. Choose 1 answer: Choose 1 answer: (Choice A) A $ \int_0^2 \int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}} dy \, dx$ (Choice B) B $ \int_0^2 \int_{-\sqrt{1 - (x-1)^2}}^{\sqrt{1 - (x-1)^2}} dy \, dx$ (Choice C) C $ \int_0^2 \int_{-\sqrt{1 - x^2}}^{\sqrt{1 - x^2}} dy \, dx$ (Choice D) D $ \int_0^2 \int_{-\sqrt{4 - (x-1)^2}}^{\sqrt{4 - (x-1)^2}} dy \, dx$
Explanation: The first step whenever we want to switch bounds is to sketch the region of integration that we're given. Here, we see $1 - \sqrt{1 - y^2} < x < 1 + \sqrt{1 - y^2}$ and $-1 < y < 1$. Therefore: ${0.5}$ ${1}$ ${1.5}$ ${2}$ ${2.5}$ ${\llap{-}0.5}$ ${0.5}$ ${1}$ ${1.5}$ ${\llap{-}0.5}$ ${\llap{-}1}$ ${\llap{-}1.5}$ $y$ $x$ Because we're switching bounds to $dy \, dx$, we need to start with numeric bounds for $x$. We see that $0 < x < 2$. Now we can define $y$ in terms of $x$. $-\sqrt{1 - (x - 1)^2} < y < \sqrt{1 - (x - 1)^2}$ We want to pay attention especially to how this $y$ bound works at the edge of the $x$ bound. For example, at $x = 0$, the $y$ bound makes $y = 0$ as expected. In conclusion, the double integral after switching bounds is: $ \int_0^2 \int_{-\sqrt{1 - (x-1)^2}}^{\sqrt{1 - (x-1)^2}} dy \, dx$